# Convergence of Taylor Series of Function Analytic on Disk/Lemma

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## Lemma

Let $y > 1$.

Then:

- $\ds \lim_{n \mathop \to \infty} \frac n {y^n} = 0$

## Proof

Note that $\ln y > 0$ as $y > 1$.

\(\ds \lim_{n \mathop \to \infty} \frac n {y^n}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac n {\paren {e^{\ln y} }^n}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac n {e^{\paren {\ln y} n} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 0\) | as $\ds \lim_{x \mathop \to \infty} \frac x {e^{\paren {\ln y} x} } = 0$ by Limit at Infinity of Polynomial over Complex Exponential as $\ln y > 0$ |

$\blacksquare$